Chapter 8.7, Problem 1E

Interpretation Introduction

Interpretation:

The integral ${\displaystyle \underset{{\text{r}}_{\text{0}}}{\overset{{\text{r}}_{\text{1}}}{\int }}\frac{\text{dr}}{\text{r}\left({\text{1-r}}^{\text{2}}\right)}}$ is to be evaluated using partial fractions as well as ${\text{r}}_{\text{1}}\text{ = }{\left({\text{1+-e}}^{\text{-4π}}\left({\text{r}}_{\text{0}}^{\text{-2}}\text{-1}\right)\right)}^{\frac{\text{-1}}{\text{2}}}$ and $\text{P'}\left(r^{*}\right){\text{ = e}}^{\text{-4π}}$ is to be proved.

Concept Introduction:

Poincare map is defined by ${\text{x}}_{\text{k+1}}\text{ = P}\left({\text{x}}_{\text{k}}\right)$.

Answer to Problem 1E

Solution:

The integral ${\displaystyle \underset{{\text{r}}_{\text{0}}}{\overset{{\text{r}}_{\text{1}}}{\int }}\frac{\text{dr}}{\text{r}\left({\text{1-r}}^{\text{2}}\right)}}$ is evaluated using partial fractions as ${\text{r}}_{\text{1}}\text{ = }{\left({\text{1+-e}}^{\text{-4π}}\left({\text{r}}_{\text{0}}^{\text{-2}}\text{-1}\right)\right)}^{\frac{\text{-1}}{\text{2}}}$ and $\text{P'}\left(r^{*}\right){\text{ = e}}^{\text{-4π}}$ is proved below.

Explanation of Solution

Let ${\text{r}}_{\text{0}}$ be an initial condition on the surface curve S. Since $\dot{\text{θ}}\text{ = 1}$ the first return to S occurs after a time of flight $\text{t = 2π}$.

Then, according to the definition of the Poincare map

${\text{r}}_{\text{1}}\text{ = P}\left({\text{r}}_{\text{0}}\right)$

Where ${\displaystyle \underset{{\text{r}}_{\text{0}}}{\overset{{\text{r}}_{\text{1}}}{\int }}\frac{\text{1}}{\text{r}\left({\text{1-r}}^{\text{2}}\right)}\text{dr}}\text{ = }{\displaystyle \underset{\text{0}}{\overset{\text{2π}}{\int }}\text{dt}}$

${\displaystyle \underset{{\text{r}}_{\text{0}}}{\overset{{\text{r}}_{\text{1}}}{\int }}\frac{\text{1}}{\text{r}\left({\text{1-r}}^{\text{2}}\right)}\text{dr}}\text{ = 2π}$

Using partial fraction method, we can write the integral as

$\begin{array}{l}\frac{\text{1}}{\text{r}\left({\text{1-r}}^{\text{2}}\right)}\text{ = }\frac{\text{A}}{\text{r}}\text{+}\frac{\text{B}}{\text{1- r}}\text{+}\frac{\text{C}}{\text{1 + r}}\\ \text{1 = A}\left(\text{1-r}\right)\left(\text{1+r}\right)\text{+B}\left(\text{1+r}\right)\text{r+C}\left(\text{1-r}\right)\text{r}\\ \text{1 = A}\left({\text{1-r}}^{\text{2}}\right)\text{+B}\left({\text{r}}^{\text{2}}\text{+r}\right)\text{+C}\left({\text{r-r}}^{\text{2}}\right)\\ \text{1 = }\left(\text{-A+B-C}\right){\text{r}}^{\text{2}}\text{+}\left(\text{B+C}\right)\text{r+A}\end{array}$

Comparing the coefficients gives:

$\begin{array}{l}\text{-A + B - C = 0}\\ \text{B + C = 0}\\ \text{A = 1}\end{array}$

Put $\text{A = 1}$ in $\text{-A + B - C = 0}$, then $\text{B - C = 1}$

Add $\text{B - C = 1}$ and $\text{B + C = 0}$, This gives

$\text{B = }\frac{\text{1}}{\text{2}}$

Put this value in the equation $\text{B - C = 1}$. Then it gives

$\text{C = -}\frac{\text{1}}{\text{2}}$

Hence,

$\frac{\text{1}}{\text{r}\left({\text{1-r}}^{\text{2}}\right)}\text{ = }\frac{\text{1}}{\text{r}}\text{+}\frac{\text{1}}{\text{2}\left(\text{1- r}\right)}-\frac{\text{1}}{\text{2}\left(\text{1 + r}\right)}$

Insert it in the integral

$\text{2π = }{\displaystyle \underset{{\text{r}}_{\text{0}}}{\overset{{\text{r}}_{\text{1}}}{\int }}\left[\frac{\text{1}}{\text{r}}\text{+}\frac{\text{1}}{\text{2}\left(\text{1- r}\right)}-\frac{\text{1}}{\text{2}\left(\text{1 + r}\right)}\right]}\text{dr}$

$\text{2π = }{\left[\text{ln}\left|\text{r}\right|\text{ - }\frac{\text{1}}{\text{2}}\text{ln}\left|\text{r-1}\right|\text{ - }\frac{\text{1}}{\text{2}}\text{ln}\left|\text{r+1}\right|\right]}_{{\text{r}}_{\text{0}}}^{{\text{r}}_{\text{1}}}$

$\text{2π = }\left(\text{ln}\left|{\text{r}}_{\text{1}}\right|\text{ - }\frac{\text{1}}{\text{2}}\text{ln}\left|{\text{r}}_{\text{1}}\text{-1}\right|\text{ - }\frac{\text{1}}{\text{2}}\text{ln}\left|{\text{r}}_{\text{1}}\text{+1}\right|\right)\text{ - }\left(\text{ln}\left|{\text{r}}_{\text{0}}\right|\text{ - }\frac{\text{1}}{\text{2}}\text{ln}\left|{\text{r}}_{\text{0}}\text{-1}\right|\text{ - }\frac{\text{1}}{\text{2}}\text{ln}\left|{\text{r}}_{\text{0}}\text{+1}\right|\right)$

$\begin{array}{l}\text{2π = ln}\left|\frac{{\text{r}}_{\text{1}}}{\sqrt{{\text{r}}_{\text{1}}^{\text{2}}\text{-1}}}\right|\text{- ln}\left|\frac{{\text{r}}_{\text{0}}}{\sqrt{{\text{r}}_{\text{0}}^{\text{2}}\text{-1}}}\right|\\ \text{2π = }\frac{\text{ln}\left|\frac{{\text{r}}_{\text{1}}}{\sqrt{{\text{r}}_{\text{1}}^{\text{2}}\text{-1}}}\right|}{\text{ln}\left|\frac{{\text{r}}_{\text{0}}}{\sqrt{{\text{r}}_{\text{0}}^{\text{2}}\text{-1}}}\right|}\\ \\ \end{array}$

Using logarithmic identity, the above equation can be written as

$\begin{array}{l}\frac{\text{1}}{\text{2}}\text{ln}\left|\frac{{\text{r}}_{\text{1}}{\text{(r}}_{\text{0}}^{\text{2}}\text{-1)}}{{\text{r}}_{\text{0}}{\text{(r}}_{\text{1}}^{\text{2}}\text{-1)}}\right|\text{ = 2π}\\ \text{ln}\left|\frac{{\text{r}}_{\text{1}}{\text{(r}}_{\text{0}}^{\text{2}}\text{-1)}}{{\text{r}}_{\text{0}}{\text{(r}}_{\text{1}}^{\text{2}}\text{-1)}}\right|\text{ = 4π}\end{array}$

Since $\text{r = 1}$ is a limit cycle, the absolute value can be dropped. Then the above equation becomes

$\frac{{\text{r}}_{\text{1}}{\text{(r}}_{\text{0}}^{\text{2}}\text{-1)}}{{\text{r}}_{\text{0}}{\text{(r}}_{\text{1}}^{\text{2}}\text{-1)}}{\text{ = e}}^{\text{4π}}$

Rearrange it as

$\begin{array}{l}{\text{r}}_{\text{1}}^{\text{2}}\left({\text{r}}_{\text{0}}^{\text{2}}\text{ - 1}\right){\text{ = e}}^{\text{4π}}{\text{r}}_{\text{0}}^{\text{2}}\left({\text{r}}_{\text{1}}^{\text{2}}\text{ - 1}\right)\\ {\text{r}}_{\text{1}}^{\text{2}}\left({\text{r}}_{\text{0}}^{\text{2}}{\text{ - 1 - e}}^{\text{4π}}{\text{r}}_{\text{0}}^{\text{2}}\right){\text{ = - e}}^{\text{4π}}{\text{r}}_{\text{0}}^{\text{2}}\\ {\text{r}}_{\text{1}}^{\text{2}}\text{ = }\frac{{\text{e}}^{\text{4π}}{\text{r}}_{\text{0}}^{\text{2}}}{\left({\text{-r}}_{\text{0}}^{\text{2}}{\text{ + 1 + e}}^{\text{4π}}{\text{r}}_{\text{0}}^{\text{2}}\right)}\end{array}$

Divide numerator and denominator of the right-hand side expression by ${\text{e}}^{\text{4π}}{\text{r}}_{\text{0}}^{\text{2}}$

$\begin{array}{l}{\text{r}}_{\text{1}}^{\text{2}}\text{ = }\frac{\text{1}}{\left({\text{-e}}^{\text{-4π}}{\text{+ e}}^{\text{-4π}}{\text{r}}_{\text{0}}^{\text{-2}}\text{ + 1}\right)}\\ \text{ = }\frac{\text{1}}{{\text{1 + -e}}^{\text{-4π}}\left({\text{r}}_{\text{0}}^{\text{-2}}\text{-1}\right)}\end{array}$

Take the square root of both sides

$\begin{array}{l}{\text{r}}_{\text{1}}\text{ = }\sqrt{\frac{\text{1}}{{\text{1 + -e}}^{\text{-4π}}\left({\text{r}}_{\text{0}}^{\text{-2}}\text{ - 1}\right)}}\\ \text{ = }{\left({\text{1 + -e}}^{\text{-4π}}\left({\text{r}}_{\text{0}}^{\text{-2}}\text{ - 1}\right)\right)}^{\frac{\text{-1}}{\text{2}}}\end{array}$

Hence, ${\text{r}}_{\text{1}}\text{ = }{\left({\text{1 + -e}}^{\text{-4π}}\left({\text{r}}_{\text{0}}^{\text{-2}}\text{ - 1}\right)\right)}^{\frac{\text{-1}}{\text{2}}}$ is proved.

According to the definition of the Poincare map,

${\text{r}}_{\text{n+1}}\text{ = P}\left({\text{r}}_n\right)$

${\text{r}}_{\text{1}}\text{ = P}\left({\text{r}}_0\right)$

Hence,

$\text{P}\left({\text{r}}_0\right)\text{ = }{\left({\text{1+-e}}^{\text{-4π}}\left({\text{r}}_{\text{0}}^{\text{-2}}\text{-1}\right)\right)}^{\frac{\text{-1}}{\text{2}}}$

$\text{P}\left(\text{r}\right)\text{ = }{\left({\text{1+-e}}^{\text{-4π}}\left({\text{r}}^{-2}\text{-1}\right)\right)}^{\frac{\text{-1}}{\text{2}}}$

Differentiate it with respect to r

$\text{P'}\left(\text{r}\right)\text{ = }\frac{{\text{e}}^{\text{-4π}}{\text{r}}^{\text{-3}}}{{\left({\text{1+e}}^{\text{-4π}}\left({\text{r}}^{\text{-2}}\text{-1}\right)\right)}^{\frac{\text{3}}{\text{2}}}}$

Put $\text{r = 1}$. Then,

$\text{P'}\left(\text{1}\right){\text{ = e}}^{\text{-4π}}$

Hence, $\text{P'}\left(\text{1}\right){\text{ = e}}^{\text{-4π}}$ is proved.

Conclusion

A Poincare function can be written using a Poincare map definition and partial fractions method.